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re: smart whole house fans
8 may 2002
gskimin   wrote:

>i have used house fans and the biggest problem has been humidity.

we might ask, "if this seems like a good idea, why hasn't it been done
before?" one reason may be the lack of an inexpensive smart controller
with an adaptive differential thermostat function and the ability to
measure humidity. another may be low energy prices.

>i'm in michigan and, while the air cools at night, moisture stays so
>the air is the "cool clammy" type. i don't doubt i could save some money
>at the expense of some comfort, but not enough to be worthwhile for me
>here. different climate, maybe, but the dampness is not pleasant.

honeywell's $30 hih-3610-001 sensor looks like a promising way to measure
humidity, with a 0-5v output and 2% accuracy. 

nrel says a 65 f house in phila has 4954 hdd and 1104 cdd, integrated over
24 hours. maybe less, if the house can react intelligently to temperature
changes (ie control the ventilation fan) on an hourly basis. 

looking at the graph of monthly average temps, i can imagine that a
smart house might not need any heating or cooling from may through
september (or possibly april through october), which would reduce the
yearly hdd and cdd to 4788 and 17 :-) this would save about $200 with
gas at $1/therm (100k btu) and coolth with a cop of 3 (moving 3 btu
of heat with 1 btu of electricity... eer = 3.41xcop) and 15 cents/kwh.

widening the indoor comfort limits and trying to make a house cooler than
70 f on average in summertime and warmer than 70 in wintertime using
ventilation would increase the savings. we might try to make the average
house wall temp tm = 70+(70-tob)/10, where tob is the average outdoor temp
over the last 3 days. tob = 30 makes tm = 74. tob = 100 makes tm = 63.

a house with c = 4k btu/f and g = 6k btu/h-f of inside air film conductance
has an inner rc = c/g = 2/3 h, so lowering the wall temp from tm to 70 f
in 1 hour with a constant room air temp ti makes 70 = ti+(tm-ti)exp(-1.5)
= 0.777ti+0.223tm, ie ti = 90.1-0.287tm, eg ti = 69.15 for tm = 73. we
could do this calc in each hourly tmy2 simulation step and see whether
ti is inside the comfort limits, and use those limits, if needed, or use
a different ti if the ideal one cannot be obtained with full fan power. 

for the fan limit, im = (tm-ti)6k btu/h flows out of the walls into
house air, initially. with 600 kwh/mo (~3k btu/h) of internal heat
gain, im+3k = (ti-ta)(500+h), so h = ((tm-ti)6k+3k)/(ti-ta)-500
= ((73-69.15)6k+3k)/(69.15-53)-500 = 1116 cfm, if say, ta = 53 f.
this is ok, if we assume the fan can move up to 4k cfm. 

even a little ventilation will make the absolute indoor humidity the same
as the absolute outdoor humidity. moving 4k cfm of outdoor air though
a 32'x32'x16' tall the house for a mere 4 minutes replaces it all,
in a plug flow model...

here's a scheme that doesn't require measuring the indoor rh: 

      tm     to    rho  |fan   ac    furn  notes
   ---------------------|-------------------------------------------
1.    <65   |67, last resort
2.    <68   |>tm+5| low | on  | -   | -   | until tm>70, limit ti<75
3.  6872   |65
5.    >75   |>tm-5| -   | -   | on  | -   | until tm<73, last resort

we'd do even better trying to bias the average tm up in the winter
and down in the summer by venting.

here are some examples for each line of the table...

example 1. tm = 65 f. increase it to 67 f over 1 hour, using the furnace. 
67 = ti-(ti-tm)exp(-1.5) = 0.777ti+0.223tm, ie ti = 86.2-0.287tm = 67.53
if tm = 65 at the start of the hour. the house heating requirement q =
(ti-ta)500 (for conduction to the outdoors) plus (67-tm)4k (for thermal
mass charging) - 3k (for internal gain.) if ta = 20 f, q = (67.53-20)500
+67-65)4k-3k = 28767 btu for the hour, which adds to the yearly heating
bill. let's assume this never exceeds a 125k btu/h furnace capacity. 

example 2. at 70 f and 60% rh, pi = 0.449 "hg. ventilate with outdoor air at
to and rho if (to > tm+5 and po = rho/100exp(17.863-9621/(460+to))<0.449),
trying to make ti = 90.1-0.287tm. if tm = 66 f, ti = 71.2, as a goal, to
make tm = 70 after 1 hour. if to = 80 and rho = 30%, po = 0.314 "hg, and we
ventilate with h cfm, where h = ((ti-tm)6k+3k)/(ta-ti)-500 = 3386, initially.

example 2.1. if h>4k, in this calc, we limit h to 4k and use the best
constant ti we can get, over the hour, assuming it's inside the 65-75 f
limits. for instance, if ti = 70.5 f with h = 4k cfm, initially, 
tm = 70.5-(70.5-66)exp(-1.5) = 69.5 f, after 1 hour.

of course we want to use line 2 of the table in preference to line 1,
if line 2 is possible. 

example 3. if 68|---*---www---*--- 69.1        26 f --www--*--www--*--- 69.1
      ---   |         |                                    |
            |        --- 4k btu/f  ---      (thevenin     --- 4k btu/f
     1/500  |        ---           ---       equivalent)  ---
20 f --www--          |            ---                     |
                      -                                    -

rc = 8.67 h, so after an hour, tm = 26+(69.1-26)exp(-1/8.67) = 64.4 f.
too cold, and ventilation won't help, so we use line 1. making tm = 67
after 1 hour requires ti = 90.1-0.287x69.1 = 70.3 (inside the comfort
limits), which requires q = (70.3-20)500+(67-69.1)4k-3k = 13.7k btu of
heat, adding to the yearly total. tm is cooler at the end of the hour,
so the walls provided heat.

example 3.1. if tm = 69.1 f, but to = 60 f, tm = 66+(69.1-66)exp(-1/8.67)
= 68.8 f after 1 hour, so we don't use any ventilation or heat. 

lines 4 and 5 are like 1 and 2, mutatis mutandis. 

nick




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